GATE Overflow Test Series | Mixed Subjects | Test 4 | Question: 22

Question

For the Matrix $A = \begin{bmatrix} 7 & 5\\
5 & 7 \end{bmatrix}$, the normalised Eigen Vector is/are ______. (Mark all the appropriate choices)

• A. $ \begin{bmatrix} \dfrac{1}{\sqrt{2}}\\
  \dfrac{1}{\sqrt{2}} \end{bmatrix}$
• B. $\begin{bmatrix} \dfrac{-1}{\sqrt{5}}\\
  \dfrac{2}{\sqrt{5}} \end{bmatrix}$
• C. $ \begin{bmatrix} \dfrac{-1}{\sqrt{2}}\\
  \dfrac{1}{\sqrt{2}} \end{bmatrix}$
• D. $\begin{bmatrix} \dfrac{1}{\sqrt{5}}\\
  \dfrac{2}{\sqrt{5}} \end{bmatrix}$

Answer 1

The characteristic equation $\mid A - \lambda I \mid  = 0$

$\implies \begin{vmatrix} 7 - \lambda & 5 \\
5 & 7 -\lambda
\end{vmatrix}$

$\implies (7-\lambda)(7 - \lambda) - 25 = 0$

$\implies ( 7 - \lambda)^{2} - 25 = 0$

$\implies 49 + \lambda^{2} - 14 \lambda - 25 = 0$

$\implies \lambda^{2} - 14\lambda + 24 = 0$

$\implies \lambda^{2} - 12\lambda - 2\lambda + 24 = 0$

$\implies \lambda(\lambda - 12) - 2(\lambda - 12) = 0$

$\implies (\lambda - 12) (\lambda - 2) = 0$

$\implies \lambda - 12 = 0\;\text{and}\; \lambda - 2 = 0$

$\implies \lambda  = 2,12$

For $\lambda = 2,$ the eigen vector is:

We know that $AX = \lambda X$

$\implies AX - \lambda X = O$

$\implies (A - \lambda I)X = O$

$\implies (A -2I)X = O$

$\implies \begin{bmatrix} 7-2 & 5 \\ 5 & 7-2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

$\implies \begin{bmatrix} 5 & 5 \\ 5 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

Perform, the operation $R_{2} \rightarrow R_{2} - R_{1},$ we get

$\implies \begin{bmatrix} 5 & 5 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

Rank of the matrix is $1,$ and number of variables (unknowns) $n = 2.$  

Here, $r < n = 1 < 2,$ infinite number of solutions possible, we can assign $n - r = 2 - 1 = 1$ value to the variable and solve the equations.

Let $y = k$

$\implies 5x + 5y = 0$

$\implies x + y = 0$

$\implies x + k = 0$

$\implies x = -k$

$\therefore$ The eigen vector corresponding to the eigen value $\lambda = 2$ is $\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -k \\ k \end{bmatrix} = k\begin{bmatrix} -1 \\ 1 \end{bmatrix}.$

For $\lambda = 12,$ the eigen vector is:

We know that $AX = \lambda X$

$\implies AX - \lambda X = O$

$\implies (A - \lambda I)X = O$

$\implies (A -12I)X = O$

$\implies \begin{bmatrix} 7-12 & 5 \\ 5 & 7-12 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

$\implies \begin{bmatrix} -5 & 5 \\ 5 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

Perform, the operation $R_{2} \rightarrow R_{2} + R_{1},$ we get

$\implies \begin{bmatrix} -5 & 5 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

Rank of the matrix is $1,$ and number of variables (unknowns) $n = 2.$  

Here, $r < n = 1 < 2,$ infinite number of solutions possible, we can assign $n - r = 2 - 1 = 1$ value to the variable and solve the equations.

Let $y = k$

$\implies -5x + 5y = 0$

$\implies -x + y = 0$

$\implies -x + k = 0$

$\implies x = k$

$\therefore$ The eigen vector corresponding to the eigen value $\lambda = 12$ is $\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} k \\ k \end{bmatrix} = k\begin{bmatrix} 1 \\ 1 \end{bmatrix}.$

To find the normalized form of $\begin{bmatrix} a \\ b \end{bmatrix},$ we divide each element by $\sqrt{a^{2} + b^{2}}.$

Now, the normalized eigen vector of $\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} \dfrac{-1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} \end{bmatrix}$

And, the normalized eigen vector of $\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} \end{bmatrix}$

So, the correct answer is $A;C.$